It is asserted in A Course in Metric Geometry by Burago, Burago, Ivanov that

there can be no more than continuum of mutually nonisometric compact spaces

How is this proven?

Its clear that there must be at least a continuum of mutually nonisometric compact spaces, i.e. $([0,\alpha], d_{\mathbb{R}})$ for $\alpha>0$ are a family of nonisometric metric spaces, but I don't know enough set theory to have any ideas how to bound the cardinality from above. A first guess was that the fact that compact metric spaces are totally bounded should be useful?

Here are some steps - Each (infinite) compact contains a countable dense set. Order it -- $(x_1,...)$. Consider an infinite matrix $d(x_i,x_j)$. One can recover the metric $d$ from such a matrix.
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PetyaDec 10 '10 at 18:24

3 Answers
3

I think "compact" can be even weakened here to "separable and complete" (and regarding your first guess, total boundedness is essentially used to prove that compact implies separable). Here's a sketch: any such space is determined, up to isometry, by the restriction of the metric to a countable dense subset. Thus the number of such isometry classes is bounded above by the cardinality of $\mathbb{R}^{\mathbb{N}\times \mathbb{N}}$, which is the same as the cardinality of $\mathbb{R}$, which is the continuum.